Construction Materials & Management Test 1

Concept:

In case of project management the followings are definition of various terms:

Duration (D): Duration is the estimated or actual time required to complete a task or an activity.

Earliest start time (ES): It is defined as the earliest possible time at which an activity can start. It is calculated by moving from first to last event in a network diagram.

Earliest finish time (EF): It is the earliest possible time at which an activity can finish.

Relationship between earliest start time, earliest finish time and duration is given by,

Earliest finish time = Earliest start time + duration of that activity

∴ EF = ES + D

EF - ES = D

Latest start time (LS): It is defined as the latest time an activity can start and still allow the project to be completed on estimated project completion time.

Latest finish time (LF): It is the latest possible time at which an activity can finish and still allow the project to be completed on estimated project completion time.

Relationship between latest start time, latest finish time and duration is given by,

Latest finish time = Latest start time + duration of that activity

∴ LF = LS + D

Concept

 Fixed declining balance or Constant percentage method is given by 

\(FDB = 1 - {\left( {\frac{{{C_S}}}{{{C_i}}}} \right)^{\frac{1}{n}}}\)

 Where Cs is Salvage or scrap value

 Ci is the initial cost of an asset

Calculation

 Given, Cs = 1lac, Ci = 8 lac, n = 3 year

  \(\begin{array}{l} r\; = \;1 - {\left( {\frac{{{C_S}}}{{{C_I}}}} \right)^{\frac{1}{n}}}\; = \;1 - {\left( {\frac{1}{8}} \right)^{\frac{1}{3}}}\\ \; = \;1 - \frac{1}{2}\; = \;0.5 \end{array}\)

Concept

1) Normal consistency test

Consistency: Vicat’s Apparatus is used to find out the consistency, initial setting time and final setting time of the cement. In the normal consistency test we have to find out the amount of water to be added to the cement to form a cement paste of normal consistency.

Important Point:

Standard consistency is the consistency of the cement paste that permits the Vicat’s plunger of diameter 10 mm and height 50 mm to penetrate in to the mould up to depth of 33 to 35 mm from the top. 

2) Low heat Cement

This cement is manufactured by reducing the proportion of C₃A and C₃S and increasing the proportion of C₂S. This cement shows a low rate of development of strength (Due to increasing C₂S). This cement is used in mass concreting.

Example - Hydraulic structures.

3) Bogue compounds

There are four compounds (Called Bogue's Compounds) formed as a result of hydration of cement:

Bogue Compounds

Dicalcium Silicate (C₂S): This compound will undergo reaction slowly. It is responsible for the progressive strength of concrete. This is also called as Belite. 

A higher percentage of C₂S results in slow hardening, less heat of hydration, and great resistance to chemical attack.

Tricalcium silicate (C3S): This is also called as Alite. It undergoes hydration within one week and helps in the development of strength in the early stages of concrete (aka hardening).

It has the best cementitious property among all the other Bogue's Compounds. Tricalcium Silicate (C₃S) hardens rapidly and is largely responsible for the initial set and early strength.

 The cement that has more C_­­­­3S content is good for cold weather concreting.

Tricalcium aluminate (C₃A): Celite is the quickest one to react when the water is added to the cement. It is responsible for the flash setting. The increase of this content will help in the manufacture of Quick Setting Cement. 

It provides weak resistance against sulphate attack and contribution to the development of strength is significantly less than above two bogue compounds.

Tetra calcium Alumino ferrite (C₄AF): This is called as Felite. The heat of hydration is 420 J/Cal. It has the poorest cementing value but it responsible for long term gain of strength of the cement.

Steel is an alloy of iron and carbon, along with small amounts of other alloying elements or residual elements as well. The plain iron-carbon alloys (Steel) contain 0.002 - 2.1% by weight carbon. For most of the materials, it varies from 0.1-1.5%.

There are 3 types of plain carbon steel:

(i) Low-carbon steels: Carbon content in the range of < 0.3%

(ii) Medium carbon steels: Carbon content in the range of 0.3 – 0.6%.

(iii) High-carbon steels: Carbon content in the range of 0.6 – 0.8%.

Hence the carbon content (%) in mild steel (Low-carbon steel) is < 0.3%, i.e 0.15 – 0.3%

Concept:

Water Cement ratio (W/C) is the ratio of weight of water to weight of cement in a concrete mix. This ratio decides the strength and workability of concrete.

According to Abram’s law, the strength of a concrete mix is inversely related to the weight ratio of water to cement. Higher W/C ratio, lower the strength but higher is the workability.

Calculation:

\(\frac{W}{C} = 0.5 \Rightarrow C = \frac{W}{{0.5}}\)

Volume of water = 191.6 litre = 0.1916 m³

Mass of water = 0.1916 × 10³ = 191.6 kg

For 1 m³ cement content required is,

∴ \(C = \frac{{191.6}}{{0.5}} = 383.2 \;kg\)

Concept

See clause 6.2.5.1 of IS 456:2000, the strain that develops due to constant sustain loading is called creep strain but in the initial age of concrete, the creep strain of concrete is higher than later age.

As per IS 456: 2000, Clause 6.2.5.1:

The effective modulus of elasticity is given by:

\({E_{effective}} = \frac{{{E_c}}}{{1 + θ }}\)

where,

Ec = Modulus of elasticity of concrete

However, elastic strain remains constant throughout. So the creep coefficient (θ ) \(= \frac{{Creep\;strain}}{{Elastic\;strain}},\) decrease with time.

Option 1: Correct

The following table shows the composition of cement and their properties.

Option 2: Incorrect

Option 3: Incorrect

the decreasing order of rate of hydration of Portland cement compounds is

C4 AF > C3 A > C3 S > C2 S

the decreasing order of heat of hydration of Portland cement compounds is

C3 A > C3 S > C2 S > C4 AF

Option 4: Incorrect

C-H-S gel is mainly responsible for strength in cement. This is also known as Tubermorite gel. 

Formula of C-H-S gel: CaO.H₂O.SiO₂

Concept

By Straight line method 

Depreciation is given by 

\({D_m} = \frac{{{C_i} - {C_s}}}{n}\)

Where Dm is depreciation 

Ci is the Initial cost or Original cost of an asset

Cs is Salvage or scrap value (Estimated at the end of utility period)

Book value (Bm) at the end of 'm ' years life is given by 

\({B_m} = {C_i} - m × {D_m}\)

Calculation

Let the original cost (C_i) of the equipment be Rs. P

Annual depreciation (D_m) \(= \;\frac{{P - Salvage value}}{5}\)

  \(= \;\frac{{P - 1000}}{5}\)

Book value at the end of 2 year

\(\begin{array}{l} \; = \;P - 2 \times \frac{{\left( {P - 1000} \right)}}{5}\\ \therefore P - \frac{{2\left( {P - 1000} \right)}}{5}\; = \;6400 \end{array}\)

Concept:

Shrinkage in concrete: It can be defined as a decrease in either length or volume of concrete material resulting from changes in moisture content or chemical changes. Shrinkage is time-dependent and its value includes plastic shrinkage, autogenous shrinkage, drying shrinkage.

• Plastic shrinkage is contraction in volume due to water movement from the concrete while still in the plastic state or before it sets. Loss of water by evaporation from the surface of the concrete in different layers or by the absorption by aggregate or subgrade is believed to be the reasons of plastic shrinkage.
• Drying shrinkage is contraction in volume due to loss in moisture from the concrete while drying. It is the loss of capillary water and water held in gel pores that cause the change in the volume.
• Autogenous shrinkage is the change in volume due to the chemical process of hydration of cement, exclusive of effects of applied load, and change in either thermal condition or moisture content.

Thus, the cause of shrinkage in hydrated cement paste is due to the removal of moisture in the form of capillary water, interlayer water, and adsorbed water. The chemically combined water does not take part in shrinkage and cannot be removed by the process of shrinkage.

Concept

Factors Affecting workability are as follows

Concept

Certainty of project can be determined by comparing standard deviation

Standard deviation (σ ) is given by,

\(\sigma = \frac{{{t_p} - {t_o}}}{6}\)

Expected completion of time (t_E) is given by

\({t_E} = \frac{{{t_o} \;+ \;4\; \times \;{t_m} \;+ \;{t_p}}}{6}\)

Where, t_p is pessimistic time

t_m is most likely time

t_o is optimistic time

Calculation

Given,

For contractor P (t₀) = 5, (t_m) = 10, (t_P) = 13

Standard deviation \(\sigma = \frac{{{t_p} - {t_o}}}{6} = \frac{{13 - 5}}{6}\) = 1.33

Expected time (T_E) = \(\frac{{{t_o}\; + \;4\; \times\; {t_m}\; + \;{t_p}}}{6} = \frac{{5\; + \;4 \;\times \;10\; + \;13}}{6}\) = 9.67

For contractor Q (t₀) = 6, (t_m) = 9, (t_P) = 12

Standard deviation \(\sigma = \frac{{{t_p} - {t_o}}}{6} = \frac{{12 - 6}}{6}\) = 1

Expected time (T_E) = \(\frac{{{t_o}\; + \;4\; \times\; {t_m}\; +\; {t_p}}}{6} = \frac{{6\; + \;4\; \times\; 9\; + \;12}}{6}\) = 9

For contractor R (t₀) = 5, (t_m) = 10, (t_P) = 14

Standard deviation \(\sigma = \frac{{{t_p} - {t_o}}}{6} = \frac{{14 - 5}}{6}\) = 1.5

Expected time (T_E) = \(\frac{{{t_o}\; + \;4\; \times \;{t_m}\; + \;{t_p}}}{6} = \frac{{5\; + \;4\; \times \;10\; + \;14}}{6}\) = 9.833

For contractor S (t₀) = 4, (t_m) = 10, (t_P) = 13

Standard deviation \(\sigma = \frac{{{t_p} - {t_o}}}{6} = \frac{{13 - 4}}{6}\) = 1.5

Expected time (T_E) = \(\frac{{{t_o}\; + \;4\; \times\; {t_m}\; + \;{t_p}}}{6} = \frac{{4\; +\; 4 \;\times \;10\; + \;13}}{6}\) = 9.5

Since the time estimate of Contractor Q has least standard deviation (i.e. 1). So work can be completed by him in most certain manner.

For straight-line method:

Depreciation is given by:

\({\rm{D}} = \frac{{{\rm{P}} - {\rm{S}}}}{{\rm{n}}} = \frac{{12000 - 3000}}{6} = \frac{{9000}}{6} = Rs.\;1500/ - \)

For sinking fund method:

A = annual sinking fund to replace = 12000 - 3000 = Rs. 9000

\(\rm{\therefore {\rm{D}} = \frac{{9000 \times 0.04}}{{{{\left( {1 + 0.04} \right)}^6} - 1}} = Rs.{\rm{\;}}1356.50/ - }\)

Item Rate contract

Items rate contract is also known as Unit Price contract or Schedule contract. A contractor undertakes the execution of work on an item rate basis. He is required to quote rate for individual item of work on the basis of schedule of quantities (i.e., Bill of Quantities) furnished by the department. The amount to be received by the contractor, depends upon the quantities of work actually performed. 

Percentage Rate Contract

In this form of contract, the department draws up ‘item rate tender’ i.e., Bill of quantities with – rate, amount and total amount. The contractors are required to offer to carry out the work as per with the rates shown in the specific price schedule or some percentage above or some percentage below the rates indicated in the schedule of work attached with the tender.

Lump – sum Contract

In this type of contract, contractor is required to quote a fixed sum for execution of work complete in all respect in the stipulated time according to the drawing, design and specification supplied to him with the tender.

Labour contract

In labour contact, the contractor undertakes contract for the labour portion only excluding the materials which are arranged and supplied at the work site by the department / owner. The contractor engages the requisite labour and gets the work done as per drawings and specifications. It is an items rate basis for labour portion only and the contractor is paid for the quantities of work done on measurement of different items of work at the stipulated rate in the contract agreement.   

Material supply contract

In this type of contract, a contractor has to offer his rates for supply of the required quantities of materials, inclusive of all local taxes, carriage and delivery charges to the specified stores within the time limit prescribed in the tender. All such materials received should be examined and counted or measured, as the case may be, when delivery is taken. 

Cost plus percentage rate contract

Concept:

The expected completion time (t_E)

\({t_E} = \frac{{{t_0} + 4{t_L} + {t_P}}}{6}\)

Variance = \([\frac{{t_p - t_o}}{6}]^2\)

Where,

t₀ = optimistic time

t_L = most likely time

t_P = pessimistic time

The probable range is from \((t_e - 3\sigma)\; to\;(t_e + 3\sigma)\)

Solution:

\(\begin{array}{l} {t_P}_Q = \frac{{4\; + \;4 \;\times\; 7 \;+ \;10}}{6} = 7\\ {t_{{QR}}} = \frac{{6 \;+ \;4 \;\times \;8 \;+ \;10}}{6} = 8 \end{array}\)

\(\begin{array}{l} {t_{RS}} = \frac{{9\; + \;4 \;\times\; 15 \;+ \;15}}{6} = 14\\ {t_{{ST}}} = \frac{{8 \;+ \;4 \;\times \;8 \;+ \;11}}{6} = 8.5 \end{array}\)

Standard Deviation, \(\sigma = \sqrt{2.69} = 1.64\)

The probable range is from ​\((37.5 - 3\times1.64)\; to\;(37.5 + 3\times1.64)\)

Hence, the Probable range is from 32.58 to 42.42

Calculations

 let weight of cement = x kg

 weight of sand = y kg

 weight of aggregate = z kg

 x : y : z = 1 : 2.2 : 3.6

\(\begin{array}{l} \frac{{\rm{x}}}{{\rm{y}}} = \frac{1}{{2.2}}\\ \frac{{\rm{x}}}{{\rm{z}}} = \frac{1}{{3.6}} \end{array}\)

Volume of concrete = V_w + V_s

\(0.40{\rm{\;}} = {\rm{\;}}\frac{{{\rm{weight\;of\;water}}}}{{{\rm{Density\;of\;water}}}} + \frac{{{\rm{weight\;of\;solids}}}}{{{\rm{Density\;of\;solids}}}}\)

\({\rm{water\;cement\;ratio}} = \frac{{{\rm{weight\;of\;water}}}}{{{\rm{weight\;of\;cement}}}}\)

0.55 x = W_W

_{\(0.40 = \frac{{0.55{\rm{x}}}}{{1000}} + \left[ {\frac{{\rm{x}}}{{{{\rm{G}}_{\rm{c}}} \times 1000}}} \right] + \frac{{\rm{y}}}{{{{\rm{G}}_{\rm{s}}} \times 1000}} + \frac{{\rm{z}}}{{{{\rm{G}}_{\rm{a}}} \times 1000}}\)}

\(400{\rm{\;}} = {\rm{\;}}0.55{\rm{x\;}} + {\rm{\;}}\frac{{\rm{x}}}{{{{\rm{G}}_{\rm{c}}}}} + \frac{{\rm{y}}}{{{{\rm{G}}_{\rm{s}}}}} + \frac{{\rm{z}}}{{{{\rm{G}}_{\rm{a}}}}}\)

G_c = 3.15

G_s = 2.65

G_a = 2.7

y = 2.2 x

z = 3.6 x

\(\begin{array}{l} 400 = 0.55{\rm{x}} + \frac{{\rm{x}}}{{3.15}} + \frac{{2.2{\rm{x}}}}{{2.65}} + \frac{{3.6{\rm{x}}}}{{2.7}}\\ 400 = {\rm{x}}\left[ {\frac{0.55}{{1}}+\frac{1}{{3.15}} + \frac{{2.2}}{{2.65}} + \frac{{3.6}}{{2.7}}} \right] \end{array}\)

x = 131.97 kg